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On Monday, a person mailed 8 packages weighing an average [#permalink]
13 Aug 2012, 05:57

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A

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Question Stats:

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20% (01:29) wrong based on 459 sessions

On Monday, a person mailed 8 packages weighing an average (arithmetic mean) of 12\frac{3}{8} pounds, and on Tuesday, 4 packages weighing an average of 15\frac{1}{4} pounds. What was the average weight, in pounds, of all the packages the person mailed on both days?

(A) 13\frac{1}{3}

(B) 13\frac{13}{16}

(C) 15\frac{1}{2}

(D) 15\frac{15}{16}

(E) 16\frac{1}{2}

Practice Questions Question: 16 Page: 154 Difficulty: 600

Re: On Monday, a person mailed 8 packages weighing an average [#permalink]
13 Aug 2012, 05:57

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SOLUTION

On Monday, a person mailed 8 packages weighing an average (arithmetic mean) of 12\frac{3}{8} pounds, and on Tuesday, 4 packages weighing an average of 15\frac{1}{4} pounds. What was the average weight, in pounds, of all the packages the person mailed on both days?

(A) 13\frac{1}{3}

(B) 13\frac{13}{16}

(C) 15\frac{1}{2}

(D) 15\frac{15}{16}

(E) 16\frac{1}{2}

The total weight of 8 packages is 8*12\frac{3}{8}=99 pounds;

The total weight of 4 packages is 4*15\frac{1}{4}=61 pounds;

The average weight of all 12 packages is \frac{total \ weight}{# \ of \ packages}=\frac{99+61}{12}=13\frac{1}{3}.

Re: On Monday, a person mailed 8 packages weighing an average [#permalink]
16 Aug 2012, 10:10

Bunuel wrote:

RESERVED FOR A SOLUTION.

Bunuel, had an off-topic request for you: Could you please post questions from non-OG sources as well? I'm not sure if that might breach a copyright arrangement bsaed on the source you use, and of course your comments on other's questions are supremely valuable for those of us subscribed to your daily updates - but if you could include occasional 700+ non-OG questions, would be much appreciated by your "followers"

_________________

How to improve your RC score, pls Kudo if helpful! http://gmatclub.com/forum/how-to-improve-my-rc-accuracy-117195.html Work experience (as of June 2012) 2.5 yrs (Currently employed) - Mckinsey & Co. (US Healthcare Analyst) 2 yrs - Advertising industry (client servicing)

Re: On Monday, a person mailed 8 packages weighing an average [#permalink]
16 Aug 2012, 23:54

Expert's post

SOLUTION

On Monday, a person mailed 8 packages weighing an average (arithmetic mean) of 12\frac{3}{8} pounds, and on Tuesday, 4 packages weighing an average of 15\frac{1}{4} pounds. What was the average weight, in pounds, of all the packages the person mailed on both days?

(A) 13\frac{1}{3}

(B) 13\frac{13}{16}

(C) 15\frac{1}{2}

(D) 15\frac{15}{16}

(E) 16\frac{1}{2}

The total weight of 8 packages is 8*12\frac{3}{8}=99 pounds;

The total weight of 4 packages is 4*15\frac{1}{4}=61 pounds;

The average weight of all 12 packages is \frac{total \ weight}{# \ of \ packages}=\frac{99+61}{12}=13\frac{1}{3}.

Re: On Monday, a person mailed 8 packages weighing an average [#permalink]
07 Sep 2012, 09:09

There must be something easy that i just don't get for me : 8*12(3/8) = 36 as you simplify the 8 between them. Therefore how do you manage to arrive at 99?

I guess it must be something different of spelling or something?

Re: On Monday, a person mailed 8 packages weighing an average [#permalink]
07 Sep 2012, 09:15

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Gmbrox wrote:

There must be something easy that i just don't get for me : 8*12(3/8) = 36 as you simplify the 8 between them. Therefore how do you manage to arrive at 99?

I guess it must be something different of spelling or something?

Thanks a lot for your help !

It's not 12 multiplied by 3/8. it's 12\frac{3}{8}=\frac{12*8+3}{8}=\frac{99}{8} (the same way as 1\frac{1}{2}=\frac{3}{2}). _________________

Re: On Monday, a person mailed 8 packages weighing an average [#permalink]
07 Sep 2012, 09:20

Thank you a lot bunuel, Ok after reviewing the official book, i now got it, it is a mix number, it does not exists in france so that's why. If anyone has difficulties to understand like me :

Re: On Monday, a person mailed 8 packages weighing an average [#permalink]
07 Sep 2012, 10:47

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Since the final average cannot be greater than 15\frac{1}{4}, answers C, D and E are out.

We can use the property of weighted averages. 15\frac{1}{4}=15\frac{2}{8}, the distance between the two initial averages is almost 3. Since the number of packages are in a ratio of 8:4 = 2:1, the differences between the final average and the initial averages are in a ratio 1:2. So, the distance between 12\frac{3}{8} and the final average is almost 1, close to 12\frac{3}{8}+1\approx{13}\frac{1}{4}. The final answer should be close to 13\frac{1}{4}.

Answer A. _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: On Monday, a person mailed 8 packages weighing an average [#permalink]
26 Oct 2013, 03:38

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Re: On Monday, a person mailed 8 packages weighing an average [#permalink]
15 Nov 2014, 01:55

I used ratio of packages, which is 2:1. converted both to the same fractions, so 15 1/4 = 15 2/8

2x(12 3/8) + 1x( 15 2/8) = 24+15+ 6/8+2/8 = 39 and 8/8, 8/8 is also obviously 1. Could also be together 40 but that's not easily divisible with three and you know you're left with a remainder. Instead just: 39/3 + 1/3 = 13 and 1/3

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